# How do you differentiate  f(x)= sin(x-2)^3  using the chain rule.?

Feb 2, 2016

$f ' \left(x\right) = 3 {\left(x - 2\right)}^{2} \cos \left[{\left(x - 2\right)}^{3}\right]$

#### Explanation:

The chain rule states that if $y = f \left(u\right) \mathmr{and} u = f \left(x\right)$ then
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

Furthermore, the power rule states that
$\frac{d}{\mathrm{dx}} {\left[u \left(x\right)\right]}^{n} = n \cdot {\left[u \left(x\right)\right]}^{n - 1} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

And then finally we also have that $\frac{d}{\mathrm{dx}} \sin \left[u \left(x\right)\right] = \cos u \cdot \frac{\mathrm{du}}{\mathrm{dx}}$.

So application of these rules yields the following result :

$\frac{d}{\mathrm{dx}} \left[\sin {\left(x - 2\right)}^{3}\right] = \cos \left[{\left(x - 2\right)}^{3}\right] \cdot 3 {\left(x - 2\right)}^{2} \cdot 1$

$= 3 {\left(x - 2\right)}^{2} \cos \left[{\left(x - 2\right)}^{3}\right]$